GSEB Class 10 Maths Solutions Chapter 5 Arithmetic Progressions Exercise 5.3

Question 1. Find the sum of the following APs
(i) 2, 7, 12, ... to 10 terms
(ii) – 37, – 33, – 29, ... to 12 terms
(iii) 0.6, 1.7, 2.8, ... to 100 terms
(iv) \( \frac { 1 }{15} \), \( \frac { 1 }{ 12 } \), \( \frac { 1 }{ 10 } \) to 11 terms.
Answer:
(i) 2, 7, 12, ... to 10 terms
Here, the first term \( a = 2 \), and the common difference \( d = 7 - 2 = 5 \).
We know that the sum of an AP, \( S_n = \frac { n }{ 2 } [2a + (n - 1)d] \).
For 10 terms, \( n = 10 \).
\( S_{10} = \frac {10}{ 2 } [2 \times 2 + (10 - 1) \times 5 ] \)
\( S_{10} = 5 [4 + 9 \times 5] \)
\( S_{10} = 5 [4 + 45] \)
\( S_{10} = 5 \times 49 \)
\( S_{10} = 245 \)
Therefore, the sum of the first 10 terms of this AP is 245.
In simple words: To find the sum, we identified the first term, common difference, and number of terms. Then we used the formula for the sum of an arithmetic progression to calculate the total.

Exam Tip: Remember to clearly identify 'a', 'd', and 'n' before applying the sum formula. Double-check your calculations, especially with multiplication and addition steps.

(ii) – 37, – 33, – 29, ... to 12 terms
Answer:
Here, the first term \( a = -37 \), and the common difference \( d = -33 - (-37) = -33 + 37 = 4 \).
We know that the sum of an AP, \( S_n = \frac { n }{ 2 } [2a + (n - 1)d] \).
For 12 terms, \( n = 12 \).
\( S_{12} = \frac {12}{ 2 } [2 \times (-37) + (12 - 1) \times 4] \)
\( S_{12} = 6 [-74 + 11 \times 4] \)
\( S_{12} = 6 [-74 + 44] \)
\( S_{12} = 6 \times (-30) \)
\( S_{12} = -180 \)
Therefore, the sum of the first 12 terms of this AP is -180.
In simple words: We found the starting number, the gap between numbers, and how many numbers to add. Then we used the sum formula to get the total.

Exam Tip: Be very careful with negative numbers when calculating the common difference and substituting into the formula. A common mistake is mismanaging the signs.

(iii) 0.6, 1.7, 2.8, ... to 100 terms
Answer:
Here, the first term \( a = 0.6 \), and the common difference \( d = 1.7 - 0.6 = 1.1 \).
We know that the sum of an AP, \( S_n = \frac { n }{ 2 } [2a + (n - 1)d] \).
For 100 terms, \( n = 100 \).
\( S_{100} = \frac { 100 }{ 2 } [2 \times 0.6 + (100 - 1) \times 1.1] \)
\( S_{100} = 50 [1.2 + 99 \times 1.1] \)
\( S_{100} = 50 [1.2 + 108.9] \)
\( S_{100} = 50 [110.1] \)
\( S_{100} = 5505 \)
Therefore, the sum of the first 100 terms of this AP is 5505.
In simple words: We used the given starting term, the constant difference, and the total count of terms. With these, we applied the sum formula for an AP to get the overall total.

Exam Tip: When working with decimals, maintain precision in your calculations to avoid rounding errors in the final sum. It's often helpful to align decimal points for addition and subtraction.

(iv) \( \frac { 1 }{ 15 } \), \( \frac { 1 }{ 12 } \), \( \frac { 1 }{ 10 } \), .......to 11 terms
Answer:
Here, the first term \( a = \frac { 1 }{ 15 } \).
The common difference \( d = \frac { 1 }{ 12 } - \frac { 1 }{ 15 } = \frac { 5 - 4 }{ 60 } = \frac { 1 }{ 60 } \).
For 11 terms, \( n = 11 \).
We know that the sum of an AP, \( S_n = \frac { n }{ 2 } [2a + (n - 1)d] \).
\( S_{11} = \frac { 11 }{ 2 } \left[ 2 \times \frac { 1 }{ 15 } + (11 - 1) \times \frac { 1 }{ 60 } \right] \)
\( S_{11} = \frac { 11 }{ 2 } \left[ \frac { 2 }{ 15 } + 10 \times \frac { 1 }{ 60 } \right] \)
\( S_{11} = \frac { 11 }{ 2 } \left[ \frac { 2 }{ 15 } + \frac { 1 }{ 6 } \right] \)
To add the fractions, find a common denominator, which is 30.
\( S_{11} = \frac { 11 }{ 2 } \left[ \frac { 4 }{ 30 } + \frac { 5 }{ 30 } \right] \)
\( S_{11} = \frac { 11 }{ 2 } \times \frac { 9 }{ 30 } \)
\( S_{11} = \frac { 11 }{ 2 } \times \frac { 3 }{ 10 } \)
\( S_{11} = \frac { 33 }{ 20 } \)
Therefore, the sum of the first 11 terms of this AP is \( \frac { 33 }{ 20 } \).
In simple words: We calculated the starting fraction and the common fractional difference. Then, we put these values into the sum formula for an arithmetic progression and simplified the fractions to find the final sum.

Exam Tip: When dealing with fractions in an AP, always find a common denominator for 'a' and 'd' to perform calculations accurately. Simplify fractions at each step to make the process easier.

Question 2. Find the sums given below:
(i) 7 + 10\( \frac {1}{2 } \) + 14 + ......... + 84
Answer:
Here, the first term \( a = 7 \). The last term \( l = 84 \).
The common difference \( d = 10\frac {1}{2} - 7 = \frac { 21 }{ 2 } - 7 = \frac { 21 - 14 }{ 2 } = \frac { 7 }{ 2 } \).
We know that the \( n^{th} \) term of an AP is \( a_n = a + (n - 1)d \).
Since \( a_n = l \), we have:
\( 84 = 7 + (n - 1) \frac { 7 }{ 2 } \)
\( 84 - 7 = (n - 1) \frac { 7 }{ 2 } \)
\( 77 = (n - 1) \frac { 7 }{ 2 } \)
\( (n - 1) = \frac { 77 \times 2 }{ 7 } \)
\( (n - 1) = 11 \times 2 \)
\( n - 1 = 22 \)
\( n = 23 \)
Now, we find the sum of the AP using \( S_n = \frac { n }{ 2 } [a + l] \).
\( S_{23} = \frac { 23 }{ 2 } [7 + 84] \)
\( S_{23} = \frac { 23 }{ 2 } [91] \)
\( S_{23} = \frac { 2093 }{ 2 } \)
\( S_{23} = 1046\frac { 1 }{ 2 } \)
Therefore, the required sum is \( 1046\frac { 1 }{ 2 } \).
In simple words: We found the first and last numbers, and the difference between them. We used these to calculate how many numbers are in the list. Finally, we added all these numbers together using the sum formula.

Exam Tip: When given the first and last terms, first calculate 'n' using the \( a_n \) formula, then use the simpler sum formula \( S_n = \frac { n }{ 2 } [a + l] \). This reduces calculation steps.

(ii) 34 + 32 + 30 + ......... + 10
Answer:
Here, the first term \( a = 34 \), and the last term \( l = 10 \).
The common difference \( d = 32 - 34 = -2 \).
Let the number of terms in the AP be \( n \).
We know that \( l = a + (n - 1)d \).
\( 10 = 34 + (n - 1) \times (-2) \)
\( 10 - 34 = (n - 1) \times (-2) \)
\( -24 = (n - 1) \times (-2) \)
\( \frac { -24 }{ -2 } = n - 1 \)
\( 12 = n - 1 \)
\( n = 13 \)
Now, we find the sum of the AP using \( S_n = \frac { n }{ 2 } [a + l] \).
\( S_{13} = \frac { 13 }{ 2 } [34 + 10] \)
\( S_{13} = \frac { 13 }{ 2 } [44] \)
\( S_{13} = 13 \times 22 \)
\( S_{13} = 286 \)
Therefore, the required sum is 286.
In simple words: We figured out the first and last numbers, and how much they change by. This helped us count the total numbers. Then, we added all the numbers in the series.

Exam Tip: Pay attention to the common difference 'd' when it's negative. Ensure that subtraction and division with negative numbers are handled correctly. Always find 'n' first if the last term is given.

(iii) – 5 + (-8) + (-11) + ......... + (-230)
Answer:
Here, the first term \( a = -5 \), and the last term \( l = -230 \).
The common difference \( d = -8 - (-5) = -8 + 5 = -3 \).
Let the number of terms in the AP be \( n \).
We know that \( l = a + (n - 1)d \).
\( -230 = -5 + (n - 1) \times (-3) \)
\( -230 + 5 = (n - 1) \times (-3) \)
\( -225 = (n - 1) \times (-3) \)
\( n - 1 = \frac { -225 }{ -3 } \)
\( n - 1 = 75 \)
\( n = 76 \)
Now, we find the sum of the AP using \( S_n = \frac { n }{ 2 } [a + l] \).
\( S_{76} = \frac { 76 }{ 2 } [-5 + (-230)] \)
\( S_{76} = 38 [-5 - 230] \)
\( S_{76} = 38 [-235] \)
\( S_{76} = -8930 \)
Therefore, the required sum is -8930.
In simple words: We found the first and last terms, and the negative common difference. We used these to count how many terms there are. Finally, we calculated the total sum of all these terms.

Exam Tip: Handling multiple negative numbers requires extra care. Write down each step clearly to minimize errors, especially when subtracting or multiplying negative values.

Question 3. In an AP.
(i) Given a = 5, d = 3, \( a_n = 50 \), find n and \( S_n \).
Answer:
We are given the first term \( a = 5 \), common difference \( d = 3 \), and the \( n^{th} \) term \( a_n = 50 \).
We know that \( a_n = a + (n - 1)d \).
\( 50 = 5 + (n - 1) \times 3 \)
\( 50 - 5 = (n - 1) \times 3 \)
\( 45 = (n - 1) \times 3 \)
\( n - 1 = \frac { 45 }{ 3 } \)
\( n - 1 = 15 \)
\( n = 15 + 1 \)
\( n = 16 \)
Now, we find the sum of the AP using \( S_n = \frac { n }{ 2 } [a + a_n] \).
\( S_{16} = \frac { 16 }{ 2 } [5 + 50] \)
\( S_{16} = 8 [55] \)
\( S_{16} = 8 \times 55 \)
\( S_{16} = 440 \)
Therefore, \( n = 16 \) and \( S_n = 440 \).
In simple words: We used the given first term, common difference, and the last term to find the total count of terms. Then we used this count, along with the first and last terms, to calculate the sum of all terms.

Exam Tip: When both \( a_n \) and 'n' are unknown initially, first use the \( a_n \) formula to determine 'n', and then proceed to find \( S_n \). This method is usually more straightforward.

(ii) Given a = 7, \( a_{13} = 35 \), find d and \( S_{13} \).
Answer:
We are given the first term \( a = 7 \) and the \( 13^{th} \) term \( a_{13} = 35 \).
We know that \( a_n = a + (n - 1)d \).
So, for \( n = 13 \):
\( a_{13} = a + (13 - 1)d \)
\( 35 = 7 + 12d \)
\( 35 - 7 = 12d \)
\( 28 = 12d \)
\( d = \frac { 28 }{ 12 } \)
\( d = \frac { 7 }{ 3 } \)
Now, we find the sum of the first 13 terms using \( S_n = \frac { n }{ 2 } [a + a_n] \).
\( S_{13} = \frac { 13 }{ 2 } [7 + 35] \)
\( S_{13} = \frac { 13 }{ 2 } [42] \)
\( S_{13} = 13 \times 21 \)
\( S_{13} = 273 \)
Therefore, \( d = \frac { 7 }{ 3 } \) and \( S_{13} = 273 \).
In simple words: We used the first term and the 13th term to figure out the constant difference between terms. Then, using the first and 13th terms, we found the sum of the first 13 terms.

Exam Tip: When you have the first term and a specific \( n^{th} \) term, you can directly find the common difference 'd'. Then use the sum formula for 'n' terms with the first and last terms.

(iii) Given \( a_{12} = 37 \), d = 3, find a and \( S_{12} \).
Answer:
We are given the \( 12^{th} \) term \( a_{12} = 37 \) and common difference \( d = 3 \).
We know that \( a_n = a + (n - 1)d \).
For \( n = 12 \):
\( a_{12} = a + (12 - 1)d \)
\( 37 = a + 11 \times 3 \)
\( 37 = a + 33 \)
\( a = 37 - 33 \)
\( a = 4 \)
Now, we find the sum of the first 12 terms using \( S_n = \frac { n }{ 2 } [a + a_n] \).
\( S_{12} = \frac { 12 }{ 2 } [4 + 37] \)
\( S_{12} = 6 [41] \)
\( S_{12} = 6 \times 41 \)
\( S_{12} = 246 \)
Therefore, \( a = 4 \) and \( S_{12} = 246 \).
In simple words: We used the 12th term and the constant difference to find the starting term. After that, we calculated the total sum of the first 12 terms using the first and last terms.

Exam Tip: If you are given a specific term and the common difference, work backwards using the \( a_n \) formula to find the first term 'a' before calculating the sum.

(iv) Given \( a_3 = 15 \), \( S_{10} = 125 \), find d and \( a_{10} \).
Answer:
We are given the third term \( a_3 = 15 \) and the sum of the first 10 terms \( S_{10} = 125 \).
From \( a_3 = 15 \), we know \( a + (3 - 1)d = 15 \), so:
(1) \( a + 2d = 15 \)
From \( S_{10} = 125 \), we know \( S_n = \frac { n }{ 2 } [2a + (n - 1)d] \).
\( 125 = \frac { 10 }{ 2 } [2a + (10 - 1)d] \)
\( 125 = 5 [2a + 9d] \)
\( \frac { 125 }{ 5 } = 2a + 9d \)
(2) \( 25 = 2a + 9d \)
Now, we have a system of two linear equations:
1. \( a + 2d = 15 \)
2. \( 2a + 9d = 25 \)
Multiply equation (1) by 2:
\( 2(a + 2d) = 2 \times 15 \)
\( 2a + 4d = 30 \)
Subtract this new equation from equation (2):
\( (2a + 9d) - (2a + 4d) = 25 - 30 \)
\( 5d = -5 \)
\( d = -1 \)
Substitute \( d = -1 \) back into equation (1):
\( a + 2(-1) = 15 \)
\( a - 2 = 15 \)
\( a = 15 + 2 \)
\( a = 17 \)
Now, we need to find \( a_{10} \).
\( a_{10} = a + (10 - 1)d \)
\( a_{10} = a + 9d \)
\( a_{10} = 17 + 9 \times (-1) \)
\( a_{10} = 17 - 9 \)
\( a_{10} = 8 \)
Therefore, \( d = -1 \) and \( a_{10} = 8 \).
In simple words: We used the third term to make one equation and the sum of the first 10 terms to make another. We solved these two equations together to find the common difference and the first term. Finally, we used these values to find the 10th term.

Exam Tip: When given two different terms or sums, set up a system of linear equations to solve for 'a' and 'd'. Always check your calculations after solving the system to ensure accuracy.

(v) Given d = 5, \( S_9 = 75 \), find a and \( a_9 \).
Answer:
We are given the common difference \( d = 5 \) and the sum of the first 9 terms \( S_9 = 75 \).
We know that \( S_n = \frac { n }{ 2 } [2a + (n - 1)d] \).
For \( n = 9 \):
\( S_9 = \frac { 9 }{ 2 } [2a + (9 - 1)d] \)
\( 75 = \frac { 9 }{ 2 } [2a + 8 \times 5] \)
\( 75 = \frac { 9 }{ 2 } [2a + 40] \)
Multiply both sides by 2 and divide by 9:
\( \frac { 75 \times 2 }{ 9 } = 2a + 40 \)
\( \frac { 150 }{ 9 } = 2a + 40 \)
\( \frac { 50 }{ 3 } = 2a + 40 \)
\( 2a = \frac { 50 }{ 3 } - 40 \)
\( 2a = \frac { 50 - 120 }{ 3 } \)
\( 2a = \frac { -70 }{ 3 } \)
\( a = \frac { -70 }{ 3 \times 2 } \)
\( a = \frac { -35 }{ 3 } \)
Now, we find \( a_9 \).
\( a_9 = a + (9 - 1)d \)
\( a_9 = a + 8d \)
\( a_9 = \frac { -35 }{ 3 } + 8 \times 5 \)
\( a_9 = \frac { -35 }{ 3 } + 40 \)
\( a_9 = \frac { -35 + 120 }{ 3 } \)
\( a_9 = \frac { 85 }{ 3 } \)
Therefore, \( a = \frac { -35 }{ 3 } \) and \( a_9 = \frac { 85 }{ 3 } \).
In simple words: We used the common difference and the sum of the first nine terms to find the first term. Then, using this first term and the common difference, we calculated the ninth term.

Exam Tip: When dealing with fractions in calculations, ensure you handle common denominators correctly during addition and subtraction. Simplify the fractional results for 'a' and 'd' before using them further.

(vi) Given a = 2, d = 8, \( S_n = 90 \), find n and \( a_n \).
Answer:
We are given the first term \( a = 2 \), common difference \( d = 8 \), and sum of \( n \) terms \( S_n = 90 \).
We know that \( S_n = \frac { n }{ 2 } [2a + (n - 1)d] \).
\( 90 = \frac { n }{ 2 } [2 \times 2 + (n - 1) \times 8] \)
\( 180 = n [4 + 8n - 8] \)
\( 180 = n [8n - 4] \)
\( 180 = 4n(2n - 1) \)
Divide by 4:
\( 45 = n(2n - 1) \)
\( 45 = 2n^2 - n \)
\( 2n^2 - n - 45 = 0 \)
We solve this quadratic equation by factoring:
\( 2n^2 - 10n + 9n - 45 = 0 \)
\( 2n(n - 5) + 9(n - 5) = 0 \)
\( (n - 5)(2n + 9) = 0 \)
This gives two possible values for \( n \): \( n = 5 \) or \( n = -\frac { 9 }{ 2 } \).
Since \( n \) represents the number of terms, it must be a natural number.
Therefore, \( n = 5 \).
Now, we find \( a_n \) (which is \( a_5 \)).
\( a_5 = a + (5 - 1)d \)
\( a_5 = 2 + 4 \times 8 \)
\( a_5 = 2 + 32 \)
\( a_5 = 34 \)
Therefore, \( n = 5 \) and \( a_n = 34 \).
In simple words: We used the first term, common difference, and the total sum to set up a quadratic equation for the number of terms. After solving, we found the valid number of terms and then used it to calculate the last term in the series.

Exam Tip: When solving for 'n' which results in a quadratic equation, always remember that 'n' must be a positive integer (natural number). Reject any negative or fractional solutions for 'n'.

(vii) Given a = 8, \( a_n = 62 \), \( S_n = 210 \), find n and d.
Answer:
We are given the first term \( a = 8 \), the \( n^{th} \) term \( a_n = 62 \), and the sum of \( n \) terms \( S_n = 210 \).
We know that \( S_n = \frac { n }{ 2 } [a + a_n] \).
\( 210 = \frac { n }{ 2 } [8 + 62] \)
\( 210 = \frac { n }{ 2 } [70] \)
\( 210 = 35n \)
\( n = \frac { 210 }{ 35 } \)
\( n = 6 \)
Now, we find 'd' using the formula \( a_n = a + (n - 1)d \).
\( 62 = 8 + (6 - 1)d \)
\( 62 = 8 + 5d \)
\( 62 - 8 = 5d \)
\( 54 = 5d \)
\( d = \frac { 54 }{ 5 } \)
Therefore, \( n = 6 \) and \( d = \frac { 54 }{ 5 } \).
In simple words: We used the first term, last term, and total sum to quickly find how many terms are in the sequence. Then, with the number of terms and the first and last terms, we calculated the constant difference.

Exam Tip: If you have 'a', \( a_n \), and \( S_n \), always prioritize using \( S_n = \frac { n }{ 2 } [a + a_n] \) to find 'n' first, as it's typically the most direct path to 'n'.

(viii) Given \( a_n = 4 \), d = 2, \( S_n = -14 \), find n and a.
Answer:
We are given the \( n^{th} \) term \( a_n = 4 \), common difference \( d = 2 \), and sum of \( n \) terms \( S_n = -14 \).
We know that \( a_n = a + (n - 1)d \).
\( 4 = a + (n - 1)2 \)
\( 4 = a + 2n - 2 \)
(1) \( a = 6 - 2n \)
We also know that \( S_n = \frac { n }{ 2 } [a + a_n] \).
\( -14 = \frac { n }{ 2 } [a + 4] \)
Substitute \( a = 6 - 2n \) from equation (1) into this equation:
\( -14 = \frac { n }{ 2 } [(6 - 2n) + 4] \)
\( -14 = \frac { n }{ 2 } [10 - 2n] \)
\( -28 = n [10 - 2n] \)
\( -28 = 10n - 2n^2 \)
Rearrange into a quadratic equation:
\( 2n^2 - 10n - 28 = 0 \)
Divide by 2:
\( n^2 - 5n - 14 = 0 \)
Solve the quadratic equation by factoring:
\( n^2 - 7n + 2n - 14 = 0 \)
\( n(n - 7) + 2(n - 7) = 0 \)
\( (n - 7)(n + 2) = 0 \)
This gives two possible values for \( n \): \( n = 7 \) or \( n = -2 \).
Since \( n \) represents the number of terms, it must be a positive integer.
Therefore, \( n = 7 \).
Substitute \( n = 7 \) back into equation (1) to find 'a':
\( a = 6 - 2(7) \)
\( a = 6 - 14 \)
\( a = -8 \)
Therefore, \( n = 7 \) and \( a = -8 \).
In simple words: We used the last term and common difference to write the first term in terms of 'n'. Then we substituted this into the sum formula. This gave us a quadratic equation for 'n', which we solved. Finally, we used 'n' to find the first term 'a'.

Exam Tip: When 'a' and 'n' are unknown, and you have \( a_n \), 'd', and \( S_n \), it's efficient to express 'a' in terms of 'n' from \( a_n \) formula and substitute it into the \( S_n \) formula. This leads to a quadratic in 'n'.

(ix) Given a = 3, n = 8, S = 192 find d.
Answer:
We are given the first term \( a = 3 \), number of terms \( n = 8 \), and sum of terms \( S = 192 \).
We know that \( S_n = \frac { n }{ 2 } [2a + (n - 1)d] \).
\( 192 = \frac { 8 }{ 2 } [2 \times 3 + (8 - 1)d] \)
\( 192 = 4 [6 + 7d] \)
\( \frac { 192 }{ 4 } = 6 + 7d \)
\( 48 = 6 + 7d \)
\( 48 - 6 = 7d \)
\( 42 = 7d \)
\( d = \frac { 42 }{ 7 } \)
\( d = 6 \)
Therefore, the common difference \( d = 6 \).
In simple words: We used the given first term, number of terms, and the total sum. By plugging these values into the sum formula for an AP, we could directly solve for the common difference.

Exam Tip: If 'a', 'n', and \( S_n \) are given, the \( S_n \) formula with '2a' is the most direct way to find the common difference 'd'. Carefully manage the arithmetic operations to get the correct result.

(x) Given l = 28, S = 144, and there are total 9 terms, find a.
Answer:
We are given the last term \( l = 28 \), sum of terms \( S = 144 \), and total number of terms \( n = 9 \).
We know that \( S_n = \frac { n }{ 2 } [a + l] \).
\( 144 = \frac { 9 }{ 2 } [a + 28] \)
Multiply both sides by 2 and divide by 9:
\( \frac { 144 \times 2 }{ 9 } = a + 28 \)
\( 16 \times 2 = a + 28 \)
\( 32 = a + 28 \)
\( a = 32 - 28 \)
\( a = 4 \)
Therefore, the first term \( a = 4 \).
In simple words: We had the last term, the total sum, and the number of terms. We used the simpler sum formula (which includes the first and last terms) to easily calculate the first term.

Exam Tip: When the last term 'l' (or \( a_n \)) and 'n' are given along with \( S_n \), always use the formula \( S_n = \frac { n }{ 2 } [a + l] \) as it simplifies calculations significantly by avoiding 'd'.

Question 4. How many terms of the AP: 9, 17, 25, ... must be taken to give a sum of 636?
Answer:
The given AP is 9, 17, 25, ...
Here, the first term \( a = 9 \).
The common difference \( d = 17 - 9 = 8 \).
We need to find the number of terms \( n \) such that their sum \( S_n = 636 \).
We know that \( S_n = \frac { n }{ 2 } [2a + (n - 1)d] \).
\( 636 = \frac { n }{ 2 } [2 \times 9 + (n - 1) \times 8] \)
\( 636 \times 2 = n [18 + 8n - 8] \)
\( 1272 = n [8n + 10] \)
\( 1272 = 8n^2 + 10n \)
Rearrange into a quadratic equation:
\( 8n^2 + 10n - 1272 = 0 \)
Divide by 2:
\( 4n^2 + 5n - 636 = 0 \)
We solve this quadratic equation using factoring or the quadratic formula.
Using factoring, we look for two numbers that multiply to \( 4 \times (-636) = -2544 \) and add to 5. These numbers are 53 and -48.
\( 4n^2 + 53n - 48n - 636 = 0 \)
\( n(4n + 53) - 12(4n + 53) = 0 \)
\( (4n + 53)(n - 12) = 0 \)
This gives two possible values for \( n \): \( 4n + 53 = 0 \implies n = -\frac { 53 }{ 4 } \) or \( n - 12 = 0 \implies n = 12 \).
Since \( n \) represents the number of terms, it must be a natural number.
Therefore, \( n = 12 \).
Thus, 12 terms of the AP must be taken to give a sum of 636.
In simple words: We used the first term and the difference between terms to set up an equation that includes the total sum and the number of terms. We solved this equation to find how many terms are needed for the sum to be 636.

Exam Tip: When forming a quadratic equation for 'n', simplify it by dividing by a common factor if possible. Remember to reject any non-natural number solutions for 'n' as the number of terms must be a positive integer.

Question 5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Answer:
We are given the first term \( a = 5 \), the last term \( l = 45 \), and the sum of terms \( S = 400 \).
First, we find the number of terms \( n \) using the formula \( S_n = \frac { n }{ 2 } [a + l] \).
\( 400 = \frac { n }{ 2 } [5 + 45] \)
\( 400 = \frac { n }{ 2 } [50] \)
\( 400 = 25n \)
\( n = \frac { 400 }{ 25 } \)
\( n = 16 \)
Now, we find the common difference \( d \) using the formula \( l = a + (n - 1)d \).
\( 45 = 5 + (16 - 1)d \)
\( 45 - 5 = 15d \)
\( 40 = 15d \)
\( d = \frac { 40 }{ 15 } \)
\( d = \frac { 8 }{ 3 } \)
Therefore, the number of terms is 16, and the common difference is \( \frac { 8 }{ 3 } \).
In simple words: We used the first term, last term, and the total sum to figure out how many numbers are in the sequence. Then, using this count, the first term, and the last term, we found the constant difference between each number.

Exam Tip: When 'a', 'l', and \( S_n \) are given, always find 'n' first using the \( S_n = \frac { n }{ 2 } [a + l] \) formula. This simplifies the process and allows you to find 'd' subsequently using the \( a_n \) formula efficiently.

Question 6. The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Answer:
We are given the first term \( a = 17 \), the last term \( l = 350 \), and the common difference \( d = 9 \).
First, we find the number of terms \( n \) using the formula \( l = a + (n - 1)d \).
\( 350 = 17 + (n - 1) \times 9 \)
\( 350 - 17 = (n - 1) \times 9 \)
\( 333 = (n - 1) \times 9 \)
\( n - 1 = \frac { 333 }{ 9 } \)
\( n - 1 = 37 \)
\( n = 37 + 1 \)
\( n = 38 \)
Now, we find the sum of the 38 terms using the formula \( S_n = \frac { n }{ 2 } [a + l] \).
\( S_{38} = \frac { 38 }{ 2 } [17 + 350] \)
\( S_{38} = 19 [367] \)
\( S_{38} = 19 \times 367 \)
\( S_{38} = 6973 \)
Therefore, there are 38 terms, and their sum is 6973.
In simple words: We used the first and last numbers, along with the constant difference, to count how many numbers are in the series. Then, using this count and the first and last numbers, we calculated the total sum of all the terms.

Exam Tip: When given 'a', 'l', and 'd', always calculate 'n' first using the \( l = a + (n-1)d \) formula. This 'n' can then be used with 'a' and 'l' in the simpler sum formula \( S_n = \frac{n}{2}[a+l] \).

Question 7. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Answer:
We are given the common difference \( d = 7 \) and the \( 22^{nd} \) term \( a_{22} = 149 \).
We know that \( a_n = a + (n - 1)d \).
For \( n = 22 \):
\( a_{22} = a + (22 - 1)d \)
\( 149 = a + 21 \times 7 \)
\( 149 = a + 147 \)
\( a = 149 - 147 \)
\( a = 2 \)
Now, we need to find the sum of the first 22 terms. We have \( a = 2 \), \( d = 7 \), \( n = 22 \), and \( a_{22} = 149 \).
Using the formula \( S_n = \frac { n }{ 2 } [a + a_n] \):
\( S_{22} = \frac { 22 }{ 2 } [2 + 149] \)
\( S_{22} = 11 [151] \)
\( S_{22} = 11 \times 151 \)
\( S_{22} = 1661 \)
Therefore, the sum of the first 22 terms is 1661.
In simple words: We used the constant difference and the 22nd term to figure out the starting term. Then, knowing the first and 22nd terms, we calculated the total sum of the first 22 terms.

Exam Tip: If the common difference and a specific term \( a_n \) are given, always find the first term 'a' first using \( a_n = a + (n-1)d \). Then, use the simpler sum formula \( S_n = \frac{n}{2}[a+a_n] \).

Question 8. Find the first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Answer:
Let \( a \) be the first term and \( d \) be the common difference of the AP.
We are given the second term \( a_2 = 14 \). So, \( a + (2 - 1)d = 14 \), which means:
(1) \( a + d = 14 \)
We are also given the third term \( a_3 = 18 \). So, \( a + (3 - 1)d = 18 \), which means:
(2) \( a + 2d = 18 \)
Subtract equation (1) from equation (2):
\( (a + 2d) - (a + d) = 18 - 14 \)
\( d = 4 \)
Substitute \( d = 4 \) back into equation (1):
\( a + 4 = 14 \)
\( a = 14 - 4 \)
\( a = 10 \)
Now, we need to find the sum of the first 51 terms (\( S_{51} \)). We have \( a = 10 \), \( d = 4 \), and \( n = 51 \).
Using the formula \( S_n = \frac { n }{ 2 } [2a + (n - 1)d] \):
\( S_{51} = \frac { 51 }{ 2 } [2 \times 10 + (51 - 1) \times 4] \)
\( S_{51} = \frac { 51 }{ 2 } [20 + 50 \times 4] \)
\( S_{51} = \frac { 51 }{ 2 } [20 + 200] \)
\( S_{51} = \frac { 51 }{ 2 } [220] \)
\( S_{51} = 51 \times 110 \)
\( S_{51} = 5610 \)
Therefore, the sum of the first 51 terms is 5610.
In simple words: We used the second and third terms to find the common difference and the first term. Then, with these values and the total number of terms (51), we calculated the sum of the entire sequence.

Exam Tip: If two consecutive terms of an AP are given, finding the common difference 'd' is straightforward (second term minus first term). Use these to solve for 'a' and then the desired sum.

Question 9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
Answer:
Let \( a \) be the first term and \( d \) be the common difference of the AP.
We are given that the sum of the first 7 terms \( S_7 = 49 \).
Using the formula \( S_n = \frac { n }{ 2 } [2a + (n - 1)d] \):
\( S_7 = \frac { 7 }{ 2 } [2a + (7 - 1)d] = 49 \)
\( \frac { 7 }{ 2 } [2a + 6d] = 49 \)
Multiply by \( \frac { 2 }{ 7 } \):
\( 2a + 6d = 49 \times \frac { 2 }{ 7 } \)
\( 2a + 6d = 14 \)
Divide by 2:
(1) \( a + 3d = 7 \)
We are also given that the sum of the first 17 terms \( S_{17} = 289 \).
\( S_{17} = \frac { 17 }{ 2 } [2a + (17 - 1)d] = 289 \)
\( \frac { 17 }{ 2 } [2a + 16d] = 289 \)
Multiply by \( \frac { 2 }{ 17 } \):
\( 2a + 16d = 289 \times \frac { 2 }{ 17 } \)
\( 2a + 16d = 17 \times 2 \)
\( 2a + 16d = 34 \)
Divide by 2:
(2) \( a + 8d = 17 \)
Now, we have a system of two linear equations:
1. \( a + 3d = 7 \)
2. \( a + 8d = 17 \)
Subtract equation (1) from equation (2):
\( (a + 8d) - (a + 3d) = 17 - 7 \)
\( 5d = 10 \)
\( d = 2 \)
Substitute \( d = 2 \) back into equation (1):
\( a + 3(2) = 7 \)
\( a + 6 = 7 \)
\( a = 1 \)
Finally, we need to find the sum of the first \( n \) terms (\( S_n \)). We have \( a = 1 \) and \( d = 2 \).
\( S_n = \frac { n }{ 2 } [2a + (n - 1)d] \)
\( S_n = \frac { n }{ 2 } [2 \times 1 + (n - 1) \times 2] \)
\( S_n = \frac { n }{ 2 } [2 + 2n - 2] \)
\( S_n = \frac { n }{ 2 } [2n] \)
\( S_n = n^2 \)
Therefore, the sum of the first \( n \) terms is \( n^2 \).
In simple words: We used the given sums of 7 and 17 terms to create two equations for the first term and common difference. After solving these equations to find 'a' and 'd', we used them to derive a general formula for the sum of 'n' terms.

Exam Tip: When given sums of different numbers of terms, set up simultaneous linear equations for 'a' and 'd'. Solving these first will allow you to then derive the general formula for \( S_n \).

Question 10. Show that \( a_1, a_2, a_3, ..., a_n \) form an AP where \( a_n \), is defined as below.
(i) \( a_n = 3 + 4n \)
Also find the sum of the first 15 terms in each case.
Answer:
(i) We are given \( a_n = 3 + 4n \).
To show that it forms an AP, we need to check if the common difference is constant. We will find the first few terms:
For \( n = 1 \), \( a_1 = 3 + 4 \times 1 = 7 \)
For \( n = 2 \), \( a_2 = 3 + 4 \times 2 = 11 \)
For \( n = 3 \), \( a_3 = 3 + 4 \times 3 = 15 \)
For \( n = 4 \), \( a_4 = 3 + 4 \times 4 = 19 \)
Now, calculate the differences between consecutive terms:
\( a_2 - a_1 = 11 - 7 = 4 \)
\( a_3 - a_2 = 15 - 11 = 4 \)
\( a_4 - a_3 = 19 - 15 = 4 \)
Since the difference between consecutive terms is constant (which is 4), the sequence forms an AP with first term \( a = 7 \) and common difference \( d = 4 \).
Now, we find the sum of the first 15 terms (\( S_{15} \)). We have \( a = 7 \), \( d = 4 \), and \( n = 15 \).
Using the formula \( S_n = \frac { n }{ 2 } [2a + (n - 1)d] \):
\( S_{15} = \frac { 15 }{ 2 } [2 \times 7 + (15 - 1) \times 4] \)
\( S_{15} = \frac { 15 }{ 2 } [14 + 14 \times 4] \)
\( S_{15} = \frac { 15 }{ 2 } [14 + 56] \)
\( S_{15} = \frac { 15 }{ 2 } [70] \)
\( S_{15} = 15 \times 35 \)
\( S_{15} = 525 \)
Therefore, the sum of the first 15 terms is 525.
In simple words: We checked if the difference between consecutive terms was always the same, which it was, proving it's an AP. Then we found the first term and common difference and used them to calculate the sum of the first 15 terms.

Exam Tip: To prove a sequence is an AP, calculate the difference between at least two pairs of consecutive terms (e.g., \( a_2 - a_1 \) and \( a_3 - a_2 \)). If these differences are equal, it's an AP.

(ii) \( a_n = 9 - 5n \)
Also find the sum of the first 15 terms in each case.
Answer:
(ii) We are given \( a_n = 9 - 5n \).
To show that it forms an AP, we will find the first few terms:
For \( n = 1 \), \( a_1 = 9 - 5 \times 1 = 4 \)
For \( n = 2 \), \( a_2 = 9 - 5 \times 2 = -1 \)
For \( n = 3 \), \( a_3 = 9 - 5 \times 3 = -6 \)
For \( n = 4 \), \( a_4 = 9 - 5 \times 4 = -11 \)
Now, calculate the differences between consecutive terms:
\( a_2 - a_1 = -1 - 4 = -5 \)
\( a_3 - a_2 = -6 - (-1) = -6 + 1 = -5 \)
\( a_4 - a_3 = -11 - (-6) = -11 + 6 = -5 \)
Since the difference between consecutive terms is constant (which is -5), the sequence forms an AP with first term \( a = 4 \) and common difference \( d = -5 \).
Now, we find the sum of the first 15 terms (\( S_{15} \)). We have \( a = 4 \), \( d = -5 \), and \( n = 15 \).
Using the formula \( S_n = \frac { n }{ 2 } [2a + (n - 1)d] \):
\( S_{15} = \frac { 15 }{ 2 } [2 \times 4 + (15 - 1) \times (-5)] \)
\( S_{15} = \frac { 15 }{ 2 } [8 + 14 \times (-5)] \)
\( S_{15} = \frac { 15 }{ 2 } [8 - 70] \)
\( S_{15} = \frac { 15 }{ 2 } [-62] \)
\( S_{15} = 15 \times (-31) \)
\( S_{15} = -465 \)
Therefore, the sum of the first 15 terms is -465.
In simple words: We checked if the constant difference between numbers was consistent, confirming it's an AP. Then, using the first term and the common difference, we calculated the total sum of the first 15 numbers in the series.

Exam Tip: For sequences where \( a_n \) is a linear expression in 'n', the coefficient of 'n' is the common difference 'd'. For instance, in \( a_n = 9 - 5n \), \( d = -5 \). This is a quick check, but always show the calculation of differences for full marks.

Question 11. If the sum of the first n terms of an AP is \( 4n – n^2 \), what is the first term (that is \( S_1 \))? What is the sum of first two terms? What is the second terms? Similarly, find the 3rd, the 10th and the nth terms.
Answer: We have the sum of the first n terms as \( 4n – n^2 \).
This means \( S_n = 4n – n^2 \).
If we put \( n = 1 \), we get \( S_1 = 4 \times 1 - 1^2 = 4 - 1 = 3 \).
So, the first term \( a_1 \) equals \( S_1 \), which is 3.
Thus, the first term is 3.
For \( n = 2 \), we find \( S_2 = 4 \times 2 - 2^2 = 8 - 4 = 4 \).
\( S_2 \) represents the total of the first two terms.
The second term is found by subtracting \( S_1 \) from \( S_2 \).
\( a_2 = S_2 - S_1 = 4 - 3 = 1 \).
If \( n = 3 \), we get \( S_3 = 4 \times 3 - 3^2 = 12 - 9 = 3 \).
So, the third term is \( a_3 = S_3 - S_2 = 3 - 4 = -1 \).
The common difference \( d \) is \( a_2 - a_1 = 1 - 3 = -2 \).
The 10th term \( a_{10} \) is calculated as \( a + 9d = 3 + 9 \times (-2) \).
\( a_{10} = 3 - 18 = -15 \).
The nth term \( a_n \) is equal to \( a + (n - 1)d = 3 + (n - 1)(-2) \).
\( a_n = 3 - 2n + 2 = 5 - 2n \).
In simple words: We find the sum for 1 term (\(S_1\)) and 2 terms (\(S_2\)). The first term is \(S_1\). The second term is \(S_2 - S_1\). We then use these to find the common difference and then calculate the 3rd, 10th, and nth terms.

Exam Tip: To find individual terms from the sum of n terms (\(S_n\)), remember that \(a_n = S_n - S_{n-1}\) and \(a_1 = S_1\).

Question 12. Find the sum of the first 40 positive integers divisible by 6.
Answer: The first 40 whole numbers that can be divided by 6 are 6, 12, 18, 24, and so on.
Let's check the differences between consecutive terms:
\( a_2 - a_1 = 18 - 12 = 6 \)
\( a_3 - a_2 = 24 - 18 = 6 \)
As the difference between any two consecutive terms is constant, this list of integers divisible by 6 forms an Arithmetic Progression.
In this sequence, the first term \(a\) is 6, the common difference \(d\) is 6, and the number of terms \(n\) is 40.
The total of the first 40 positive integers is \( S_{40} \).
We use the formula for the sum of an AP: \( S_n = \frac{n}{2}[2a + (n - 1)d] \).
So, \( S_{40} = \frac{40}{2} [2 \times 6 + (40 - 1) \times 6] \)
\( = 20[12 + 39 \times 6] \)
\( = 20[12 + 234] \)
\( = 20 \times 248 \)
\( = 4960 \).
In simple words: We need to add up the first 40 numbers that are multiples of 6. Since each multiple is 6 more than the last, it's an AP. We use the formula for sum of AP with \(a=6\), \(d=6\), and \(n=40\).

Exam Tip: For series of multiples, the first term \(a\) and common difference \(d\) are usually the number itself (e.g., 6 in multiples of 6).

Question 13. Find the sum of the first 15 multiples of 8.
Answer: The initial 15 numbers that are multiples of 8 are 8, 16, 24, 32, 40, and so on.
Let's look at the differences:
\( a_2 - a_1 = 16 - 8 = 8 \)
\( a_3 - a_2 = 24 - 16 = 8 \)
\( a_4 - a_3 = 32 - 24 = 8 \)
The constant difference between successive terms shows that the list of multiples of 8 forms an Arithmetic Progression.
We are provided that the first term \(a\) is 8, the common difference \(d\) is 8, and the count of terms \(n\) is 15.
The total of the first 15 multiples of 8 is \( S_{15} \).
Using the sum formula for an AP: \( S_n = \frac{n}{2}[2a + (n - 1)d] \).
\( S_{15} = \frac{15}{2} [2 \times 8 + (15 - 1) \times 8] \)
\( S_{15} = \frac{15}{2} [16 + 14 \times 8] \)
\( S_{15} = \frac{15}{2} [16 + 112] \)
\( = \frac{15}{2} [128] \)
\( = 15 \times 64 \)
\( = 960 \).
In simple words: We need to sum the first 15 numbers that can be divided evenly by 8. This creates an AP with the first term and common difference both being 8. Then we use the sum formula.

Exam Tip: Remember that the \(n\)-th multiple of a number \(k\) is \(n \times k\). In a series of multiples, the first term \(a\) and common difference \(d\) are identical to the base number.

Question 14. Find the sum of odd numbers between 0 and 50.
Answer: The odd numbers that fall between 0 and 50 are 1, 3, 5, 7, ..., 49.
Let's examine the differences:
\( a_2 - a_1 = 3 - 1 = 2 \)
\( a_3 - a_2 = 5 - 3 = 2 \)
\( a_4 - a_3 = 7 - 5 = 2 \)
Since the difference between successive terms is constant, the series of odd numbers from 0 to 50 forms an Arithmetic Progression.
We have the first term \(a\) as 1, the common difference \(d\) as 2, and the last term \(l\) as 49.
First, we need to find the number of terms \(n\). We use the formula \(l = a + (n - 1)d\).
\( 49 = 1 + (n - 1) \times 2 \)
\( 48 = (n - 1) \times 2 \)
\( n - 1 = \frac{48}{2} \)
\( n - 1 = 24 \)
\( n = 25 \).
To find the total of these odd numbers between 0 and 50, we use the sum formula:
\( S_n = \frac{n}{2}[a + l] \) (since we know the last term)
\( S_{25} = \frac{25}{2} [1 + 49] \)
\( S_{25} = \frac{25}{2} [50] \)
\( S_{25} = 25 \times 25 \)
\( = 625 \).
In simple words: We list the odd numbers from 1 to 49. This is an AP. We find how many numbers there are, then use the formula for the sum of an AP knowing the first and last terms.

Exam Tip: When finding the sum of an arithmetic progression where the last term (\(l\)) is known, you can use the formula \(S_n = \frac{n}{2}(a+l)\) after calculating \(n\).

Question 15. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the first day, Rs. 250 for the second, Rs. 300 for the third day etc, the penalty for each succeeding day being Rs. 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he delays the work by 30 days?
Answer: The penalty amount goes up by Rs. 50 each day.
The daily penalties are Rs. 200, Rs. 250, Rs. 300, and so on.
This list of penalties creates an Arithmetic Progression.
Here, the first term \(a = 200\).
The common difference \(d = 250 - 200 = 50\).
The number of days of delay \(n = 30\).
The full amount of money the contractor must pay is the sum of these penalties, \( S_{30} \).
We use the sum formula for an AP: \( S_n = \frac{n}{2}[2a + (n - 1)d] \).
\( S_{30} = \frac{30}{2} [2 \times 200 + (30 - 1) \times 50] \)
\( = 15 [400 + 29 \times 50] \)
\( = 15 \times [400 + 1450] \)
\( = 15 \times 1850 \)
\( = 27750 \) Rs.
In simple words: The penalty increases by Rs. 50 daily, forming an AP. We need to find the total penalty for 30 days, using the sum formula with \(a=200\), \(d=50\), and \(n=30\).

Exam Tip: Clearly identify the first term (\(a\)), common difference (\(d\)), and number of terms (\(n\)) from the problem statement before applying the sum formula for an AP.

Question 16. A sum of Rs. 700 is to be used to gives even cash prizes to students of a school for their overall academic performance. If each prize is Rs. 20 less than its preceding prize, find the value of each of the prizes.
Answer: As the amount of each prize is Rs. 20 less than the previous one, the amounts of the seven subsequent cash prizes form an Arithmetic Progression.
Let \(x\) represent the first prize amount.
The second prize will then be \(x - 20\).
The third prize will be \(x - 40\).
The fourth prize will be \(x - 60\).
So, the prizes are \(x, x - 20, x - 40, x - 60, \dots\).
Here, the common difference \( d = (x - 20) - x = -20 \).
The number of prizes \( n = 7 \).
The total sum of prizes \( S_7 = 700 \).
We understand that the sum of an AP is \( S_n = \frac{n}{2}[2a + (n - 1)d] \).
For 7 prizes: \( S_7 = \frac{7}{2} [2 \times x + (7 - 1) \times (-20)] \).
\( 700 = \frac{7}{2} [2x - 120] \).
\( 700 = 7 [x - 60] \).
Divide both sides by 7: \( x - 60 = \frac{700}{7} \).
\( x - 60 = 100 \).
\( x = 160 \).
The value of the first prize is Rs. 160.
The value of the second prize is \( 160 - 20 = 140 \) Rs.
The value of the third prize is \( 140 - 20 = 120 \) Rs.
The value of the fourth prize is \( 120 - 20 = 100 \) Rs.
The value of the fifth prize is \( 100 - 20 = 80 \) Rs.
The value of the sixth prize is \( 80 - 20 = 60 \) Rs.
The value of the seventh prize is \( 60 - 20 = 40 \) Rs.
In simple words: The prizes decrease by Rs. 20 each, forming an AP. We know the total sum is Rs. 700 for 7 prizes. By setting up the sum formula with the first prize as \(x\) and \(d = -20\), we solve for \(x\) and then list all prize values.

Exam Tip: When dealing with decreasing values, ensure the common difference (\(d\)) is negative. Carefully set up the sum formula and solve for the unknown first term (\(a\) or \(x\)).

Question 17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g, a section of class I will plant 1 tree, a section of class II will plant 2 trees and so on till class XII. There are three sections of each class How many trees will be planted by the students?
Answer: Class I contains three sections.
Thus, the total trees planted by Class I sections are \( 3 \times 1 = 3 \).
For Class II, the students plant \( 2 + 2 + 2 \), totaling 6 trees.
For Class III, they plant \( 3 + 3 + 3 \), totaling 9 trees.
For Class IV, they plant \( 4 + 4 + 4 \), totaling 12 trees.
Following this pattern,
Class XII students will plant \( 12 + 12 + 12 \), which is 36 trees.
We now have a sequence: 3, 6, 9, 12, ...., 36.
We notice that the quantity of trees planted by each class shows a consistent difference between consecutive classes, thereby forming an Arithmetic Progression.
Here, the first term \(a = 3\), the common difference \(d = 6 - 3 = 3\), and the number of terms \(n = 12\) (for classes I to XII).
The complete count of trees planted is the sum of this AP, \( S_{12} \).
Using the sum formula: \( S_n = \frac{n}{2}[2a + (n - 1)d] \).
\( S_{12} = \frac{12}{2} [2 \times 3 + (12 - 1) \times 3] \)
\( = 6[6 + 33] \)
\( = 6 \times 39 \)
\( = 234 \).
In simple words: Each class plants trees equal to its class number, and there are three sections per class. This makes a sequence of total trees per class (3, 6, 9...). This is an AP. We find the sum of trees planted from Class I to Class XII.

Exam Tip: Break down word problems into parts: first, find the sequence of terms, then confirm it's an AP, and finally apply the appropriate formula for the sum of an AP.

Question 18. A sapiral is made up of successive semicircles, with centres alternatively at A and B. starting with centres at A of radii 0.5 cm. 1.0cm 1.5 cm. 2.0 cm, as shown in figure, What is the total length of such a spiral made up thirteen consecutive semicircles?
Answer: The lengths of the successive semicircles, with radii \( r_1 = 0.5 \) cm, \( r_2 = 1.0 \) cm, \( r_3 = 1.5 \) cm, \( r_4 = 2.0 \) cm, are given by:
\( l_1 = \pi r_1 = 0.5\pi \)
\( l_2 = \pi r_2 = 1.0\pi \)
\( l_3 = \pi r_3 = 1.5\pi \)
\( l_4 = \pi r_4 = 2.0\pi \dots \)
Let \( a_1 = 0.5\pi \).
\( a_2 - a_1 = 1.0\pi - 0.5\pi = 0.5\pi \) cm.
\( a_3 - a_2 = 1.5\pi - 1.0\pi = 0.5\pi \) cm.
\( a_4 - a_3 = 2.0\pi - 1.5\pi = 0.5\pi \) cm.
Since the difference between any two consecutive terms remains constant, the lengths of these semicircles form an Arithmetic Progression.
In this sequence, the first term \(a\) is \(0.5\pi\), the common difference \(d\) is \(0.5\pi\), and the total number of terms \(n\) is 13.
Thus, the overall length is the sum of this AP, \( S_{13} \).
We use the sum formula for an AP: \( S_n = \frac{n}{2}[2a + (n - 1)d] \).
\( S_{13} = \frac{13}{2} [2 \times 0.5\pi + (13 - 1) \times 0.5\pi] \) cm
\( = \frac{13}{2} [\pi + 12 \times 0.5\pi] \) cm
\( = \frac{13}{2} [\pi + 6\pi] \) cm
\( = \frac{13}{2} \times 7\pi \) cm
Substitute \( \pi = \frac{22}{7} \):
\( = \frac{13 \times 7}{2} \times \frac{22}{7} \) cm
\( = 13 \times 11 \) cm
\( = 143 \) cm.
In simple words: The lengths of the semicircles form an AP because their radii increase by 0.5 cm each time. We find the first term and common difference, then use the sum formula for an AP to get the total length of 13 semicircles.

Exam Tip: The length of a semicircle is \(\pi r\). Identify the pattern of radii to determine if the lengths form an AP, then apply the sum formula.

Question 19. 200 logs are stacked in the following manner: 20 logs in the bottom row. 19 in the next row, 18 in the row next to it and so on (see figure) in how many rows are the 200 logs placed and how many logs are in the top row?
Answer: The arrangement of logs, starting from the bottom row, with subsequent rows having fewer logs, is:
20, 19, 18, ............
Let's check the differences:
\( a_2 - a_1 = 19 - 20 = -1 \)
\( a_3 - a_2 = 18 - 19 = -1 \)
As the difference between any two sequential terms is consistent, this arrangement of logs constitutes an Arithmetic Progression.
In this series, the first term \(a\) is 20, the common difference \(d\) is -1, and the total sum \(S_n\) is 200.
We use the sum formula: \( S_n = \frac{n}{2}[2a + (n - 1)d] \).
\( 200 = \frac{n}{2}[2 \times 20 + (n - 1) \times (-1)] \)
\( 400 = n[40 - n + 1] \)
\( 400 = n[41 - n] \)
\( 400 = 41n - n^2 \)
Rearranging this into a quadratic equation: \( n^2 - 41n + 400 = 0 \).
Factor the quadratic: \( n^2 - 25n - 16n + 400 = 0 \).
\( n(n - 25) - 16(n - 25) = 0 \).
\( (n - 25)(n - 16) = 0 \).
This gives two possible values for \(n\): \( n = 25 \) or \( n = 16 \).
Thus, the total count of rows can be either 25 or 16.
Now, let's find the count of logs in the nth row using \( a_n = a + (n - 1)d \).
If \( n = 25 \), then \( a_{25} = 20 + (25 - 1) \times (-1) \).
\( = 20 + 24 \times (-1) \)
\( = 20 - 24 = -4 \).
A negative number of logs is impossible, so the correct number of rows has to be 16.
Now, let's find the count of logs in the very top row (the 16th row):
\( a_{16} = a + (16 - 1)d \)
\( = 20 + 15 \times (-1) \)
\( = 20 - 15 \)
\( = 5 \) logs.
In simple words: The logs stack with one less log in each higher row, forming an AP. We use the sum formula to find the number of rows. One possible answer for rows leads to a negative number of logs, which is impossible, so the other answer is correct. Then we find the number of logs in that top row.

Exam Tip: When solving for \(n\) in quadratic equations derived from AP sum problems, always check the validity of both solutions within the context of the problem (e.g., number of items cannot be negative).

Question 20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are Len potatoes in the line (see Figure).
Answer: The distance covered to pick up the first potato is \(2 \times 5\) meters, which totals 10 meters.
For the second potato, the distance traveled is \(2 \times (5 + 3)\) meters, making it 16 meters.
The distance to retrieve the third potato is \(2 \times (5 + 3 + 3)\) meters, or 22 meters, and this pattern continues.
Let's check the differences:
\( a_2 - a_1 = 16 - 10 = 6 \) m
\( a_3 - a_2 = 22 - 16 = 6 \) m
As the difference between consecutive terms remains constant, the series of distances run forms an Arithmetic Progression.
In this sequence, the first term \(a\) is 10 m, the common difference \(d\) is 6 m, and the number of potatoes (terms) \(n\) is 10.
So, the total distance covered by the competitor is \( S_{10} \).
We use the sum formula for an AP: \( S_n = \frac{n}{2}[2a + (n - 1)d] \).
\( S_{10} = \frac{10}{2} [2 \times 10 + (10 - 1) \times 6] \)
\( S_{10} = 5 [20 + 9 \times 6] \)
\( S_{10} = 5 \times [20 + 54] \)
\( S_{10} = 5 \times 74 \)
\( S_{10} = 370 \) m.
In simple words: The runner picks up potatoes one by one, returning to the bucket each time. The distance for each potato forms an AP because the potatoes are equally spaced. We calculate the sum of these distances for 10 potatoes using the AP sum formula.

Exam Tip: In distance-based AP problems, ensure you calculate the round trip distance for each item (e.g., to the potato and back to the bucket) to correctly form the AP.